Question

Biết \(\left(\dfrac{a^2+b^2}{c^2+d^2}\right)^2=\dfrac{ab}{cd}\)với a,b,c,d \(\ne\)c\(\ne\)\(\pm\)d

Chứng minh rằng: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{a-b}{c-d}\right)^2\)

Answer 1

hình như đề sai đó bạn

Answer 2

theo bài ra ta có:

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}\\ \Rightarrow\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}\)

áp dụng tính chất dảy tỉ số bằng nhau ta có:

\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{ab}{cd}=\dfrac{2ab}{2cd}=\dfrac{a^2+b^2+2ab}{c^2+d^2+2cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{a^2+b^2-2cd}{c^2+d^2-2cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\) \(\Rightarrow\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\\ \Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{a-b}{c-d}\right)^2\left(đpcm\right)\)